| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Multiple independent time periods |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question requiring standard calculations: cumulative probabilities from tables, independence for part (b), and finding the mode by comparing consecutive probabilities in part (c). All techniques are routine for S2 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(e^{-2.4}\left(\frac{2.4^2}{2!}+\frac{2.4^3}{3!}\right)\right) = e^{-2.4}(2.88+2.304) = 0.2613+0.2090\) | M1 | Allow M1 for \(e^{-2.4}\left(\frac{2.4^2}{2!}+\frac{2.4^3}{3!}+\frac{2.4^4}{4!}\right)\). Expression must be seen. |
| \(= 0.47(0)\) | A1 | SC B1 0.47(0) with no working. |
| Answer | Marks | Guidance |
|---|---|---|
| \(1-e^{-2.4}(1+2.4)\ [=0.691558]\) | M1 | Allow one end error. Allow any \(\lambda\). |
| \((1-e^{-2.4}(1+2.4))^2\) | M1 | Squaring their probability (\(\lambda \neq 4.8\)). |
| \(= 0.478\) (3 sf) | A1 | SC B2 0.478 with no working. |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-2.4}\times\frac{2.4^r}{r!} < e^{-2.4}\times\frac{2.4^{r+1}}{(r+1)!}\) | M1 | For both expressions seen. |
| \(r+1<2.4\) | A1* | Or \(r<1.4\) (must have correct inequality). |
| Set is \(r=0,\ 1\) | DA1 | |
| 3 | For trial and error solutions: M1 for substituting one value into correct expression (can be implied by correct values). A1* for \([P(0)=0.0907]\), \(P(1)=0.218\), \(P(2)=0.261\), \(P(3)=0.209\) (accept 2sf accuracy). DA1 Set is \(r=0,1\). |
| Answer | Marks |
|---|---|
| \(r=2\) | B1 |
## Question 7(a):
$\left(e^{-2.4}\left(\frac{2.4^2}{2!}+\frac{2.4^3}{3!}\right)\right) = e^{-2.4}(2.88+2.304) = 0.2613+0.2090$ | M1 | Allow M1 for $e^{-2.4}\left(\frac{2.4^2}{2!}+\frac{2.4^3}{3!}+\frac{2.4^4}{4!}\right)$. Expression must be seen.
$= 0.47(0)$ | A1 | **SC B1** 0.47(0) with no working.
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## Question 7(b):
$1-e^{-2.4}(1+2.4)\ [=0.691558]$ | M1 | Allow one end error. Allow any $\lambda$.
$(1-e^{-2.4}(1+2.4))^2$ | M1 | Squaring their probability ($\lambda \neq 4.8$).
$= 0.478$ (3 sf) | A1 | **SC B2** 0.478 with no working.
---
## Question 7(c)(i):
$e^{-2.4}\times\frac{2.4^r}{r!} < e^{-2.4}\times\frac{2.4^{r+1}}{(r+1)!}$ | M1 | For both expressions seen.
$r+1<2.4$ | A1* | Or $r<1.4$ (must have correct inequality).
Set is $r=0,\ 1$ | DA1 |
| 3 | For trial and error solutions: M1 for substituting one value into correct expression (can be implied by correct values). A1* for $[P(0)=0.0907]$, $P(1)=0.218$, $P(2)=0.261$, $P(3)=0.209$ (accept 2sf accuracy). **DA1** Set is $r=0,1$.
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## Question 7(c)(ii):
$r=2$ | B1 |7 A random variable $X$ has the distribution $\operatorname { Po } ( 2.4 )$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( 2 \leqslant X < 4 )$.
\item Two independent values of $X$ are chosen.
Find the probability that both of these values are greater than 1 .
\item It is given that $\mathrm { P } ( X = r ) < \mathrm { P } ( X = r + 1 )$.
\begin{enumerate}[label=(\roman*)]
\item Find the set of possible values of $r$.
\item Hence find the value of $r$ for which $\mathrm { P } ( X = r )$ is greatest.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q7 [9]}}